This is not a complete answer, but I think that it might help to clear up some misunderstandings. It is *not*, in general, true that all the principal $H$-bundles over $M$ supporting a Cartan connection of type $(G,H)$ are isomorphic, though the OP's proposed example does not actually show this. I think that the following discussion may help.

To fix notation, let's recall what we mean by a "Cartan connection of type $(G,H)$": Here $G$ is a Lie group with Lie algebra $\frak{g}$ and $H$ is a Lie subgroup with Lie algebra ${\frak{h}}\subset{\frak{g}}$. The representation $\mathrm{Ad}:H\to\mathrm{Aut}({\frak{g}})$ preserves the subalgebra ${\frak{h}}$ and so induces a representation $\rho:H\to \mathrm{Aut}({\frak{g/h}})$. If $\pi:B\to M$ is a principal right $H$-bundle, let $X_v$ for $v\in\frak{h}$ be the vertical vector field on $B$ whose flow is right action by $\mathrm{exp}(tv)$. Then a *Cartan connection of type $(G,H)$* on $\pi:B\to M$ is a $\frak{g}$-valued $1$-form $\gamma:TB\to \frak{g}$ with the following properties:

- $\gamma_u:T_uB\to{\frak{g}}$ is an isomorphism for all $u\in B$.
- $\gamma\bigl(X_v(u)\bigr) = v$ for all $u\in B$ and all $v\in\frak{h}$.
- $R^*_h(\gamma) = \mathrm{Ad}(h^{-1})(\gamma)$ for all $h\in H$.

*It is important to note that not every principal right $H$-bundle over $M$ supports a Cartan connection of type $(G,H)$*. This is because such a Cartan connection $\gamma$ defines an isomorphism $\iota_\gamma:TM\to B\times_\rho {\frak{g/h}}$. To see this, let $\omega = \gamma\,\mathrm{mod}\,{\frak{h}}:TB\to {\frak{g/h}}$. The above axioms imply that $\omega_u:T_uB/V_uB\to {\frak{g/h}}$ is an isomorphism for all $u\in B$, where $V_uB\subset T_uB$ is tangent to the fiber of $\pi:B\to M$. Since we have a canonical isomorphism $T_uB/V_uB\to T_{\pi(u)}M$, it follows that we can regard $\omega$ as defining an isomorphism $\omega_u:T_{\pi(u)}M\to {\frak{g/h}}$ for all $u\in B$ that satisfies $\omega_{u\cdot h} = \rho(h^{-1})(\omega_u)$ for all $u\in B$ and all $h\in H$. By the very definition of $B\times_\rho{\frak g/h}$, this establishes the claimed isomorphism $\iota_\gamma:TM\to B\times_\rho{\frak g/h}$.

Conversely, if an isomorphism $\iota:TM\to B\times_\rho{\frak g/h}$ is given, then one can construct a Cartan connection of type $(G,H)$ on $B$.

Thus, one can see why the OPs construction starting with a line bundle $L$ endowed with a $\mathrm{U}(1)$-connection does not automatically imply that there is a Cartan connection of the desired type on $M$. For example, in this case, if a Cartan connection existed, then $TM$ would have to be isomorphic to $L\otimes \mathbb{C}^n = B\times_\rho {\frak g/h}$, and this is generally not the case.

However, there is a simpler example to demonstrate that not all $H$-bundles that admit Cartan connections of type $(G,H)$ are isomorphic: Here, let $n=3$, let $H=\mathrm{SO}(2)$ and let $G = \mathbb{R}^3\rtimes H$, where $H=\mathrm{SO}(2)$ acts on $\mathbb{R}^3$ by rotation in the second and third coordinates. An $H$-bundle $\pi:B\to M^3$ is just an $\mathrm{SO}(2)$-bundle, so it has an Euler class (which could be nonzero) and the associated bundle $B\times_\rho \mathbb{R}^3$ is a sum of a trivial bundle and a $2$-plane bundle. If there is a Cartan connection on $B$, then we get an isomorphism of $TM$ with the sum of a trivial bundle and a $2$-plane bundle. In particular, this means that $M$ is oriented and we have a nonvanishing vector field on $M$ together with a $2$-plane subbundle that has a well-defined Euler class.

Now, every oriented $3$-manifold has a trivial tangent bundle, but once one chooses a nonvanishing vector field, the Euler class of the complementary $2$-plane bundle is determined and may very well be nonzero. For example, let $M = S^1\times S^2$. If we choose the vector field tangent to the $S^1$-fibers, then the complementary $2$-plane field is nontrivial on each $S^2$-fiber. Meanwhile, if we choose a trivialization of the tangent bundle of $M$, then letting the vector field be one of the three trivializing vector fields, the complementary $2$-plane bundle will be trivial.

Thus, we can have two $H$-bundles over $M$ that are not isomorphic even though they both admit Cartan connections of type $(\mathbb{R}^3\rtimes H,\ H)$.

It follows that the very first criterion one needs to have in order for all the Cartan connections of type $(G,H)$ to have isomorphic underlying $H$-bundles is that all of the structure reductions of the canonical $\mathrm{GL}(n,\mathbb{R})$-structure on $TM$ to a $\rho(H)$-structure be isomorphic. This is a very strong condition on $\rho(H)$ and $M$, and whether it is met depends on *both* $\rho(H)$ and $M$.

Meanwhile, for most of the familiar examples in parabolic geometry, $\rho(H)$ is some large group such as $\mathrm{GL}(n,\mathbb{R})$, $\mathrm{SL}(n,\mathbb{R})$, $\mathrm{CO}(n)$, or $\mathrm{SO}(n)$, and it happens that this uniqueness is met trivially. This may account for the common (false) belief that prompted this question in the first place.